Solve ∫ 1/3x^2-1/3x+1/7 wrt x

$$\int \frac { 1 } { 3 } x ^ { 2 } - \frac { 1 } { 3 } x + \frac { 1 } { 7 } d x$$

Evaluate

$$\frac{x^{3}}{9}-\frac{x^{2}}{6}+\frac{x}{7}+С$$

Solution Steps

Integrate the sum term by term.

$$\int \frac{x^{2}}{3}\mathrm{d}x+\int -\frac{x}{3}\mathrm{d}x+\int \frac{1}{7}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\frac{\int x^{2}\mathrm{d}x}{3}-\frac{\int x\mathrm{d}x}{3}+\int \frac{1}{7}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply \frac{1}{3} times \frac{x^{3}}{3}.

$$\frac{x^{3}}{9}-\frac{\int x\mathrm{d}x}{3}+\int \frac{1}{7}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -\frac{1}{3} times \frac{x^{2}}{2}.

$$\frac{x^{3}}{9}-\frac{x^{2}}{6}+\int \frac{1}{7}\mathrm{d}x$$

Find the integral of \frac{1}{7} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{9}-\frac{x^{2}}{6}+\frac{x}{7}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{x^{3}}{9}-\frac{x^{2}}{6}+\frac{x}{7}+С$$

Differentiate w.r.t. x

$$\frac{x^{2}}{3}-\frac{x}{3}+\frac{1}{7}$$